My sense is that even though neutrons decay into a proton and an electron they are made up of quarks, it is not just some "merged" particle where, for example, the electron is orbiting the proton very closely or something (which would be basically a hydrogen atom).

Anyway, is there some way to simply shoot a stream of electrons at hydrogen ions (which I think are easy to make and are just protons) and observe, if you do it fast enough and in a great enough volume, that some neutrons, besides a great number of new hydrogen atoms, get produced?

Maybe this happens in fusion? New neutrons get produced since I recall an objection to cold fusion is that in fact no neutrons were found.

  • $\begingroup$ Not sure if it's relevant but free neutrons (neutrons not inside a nucleus) decay quite quickly and have a mean lifetime of about 14 minutes: en.wikipedia.org/wiki/Free_neutron_decay $\endgroup$ Jan 11, 2021 at 10:31
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    $\begingroup$ Just a little side-note: This is actually the mechanism behind type 2 supernovae which produces neutron stars. The resulting neutron fluid has a FAR lower degeneracy pressure for a given density, which drives a collapse of the star. $\endgroup$ Jan 12, 2021 at 11:11

5 Answers 5


The one-word answer is yes.

You are also correct that the neutron is not just a proton and electron living together. The process of merging a proton and electron proceeds via the weak force. Specifically, an up quark in the proton exchanges a W boson with the electron. The W boson carries a unit of positive charge from the quark to the electron. In that process the up quark (charge +2/3) is converted to a down quark (charge -1/3) so that the proton (uud) becomes a neutron (udd). The negatively charged electron is converted into a neutrino. This is one important point left out in your question. The full reaction is $p+e^-\to n+\nu_e$.

There is a general principle of quantum field theory called crossing symmetry that roughly states that for any process I can exchange what I call initial and final particles. So you are correct that neutron decay $n\to p+ e^- + \bar\nu_e$ implies that the process $p+e^-\to n+\nu_e$ can also happen.

This process does also happen in nature. It is one mode of radioactive decay of nuclei. Some nuclei with a sufficiently large number of protons can become more stable by absorbing one of their electrons and converting one proton into a neutron. This can happen because electron orbitals have a small but non-zero overlap with the nucleus, so that they "sometimes come into contact with" the protons.

This process can also happen artificially as you suggest. In fact, it seems that accelerators used in medical facilities produce neutrons as a by-product, exactly as you suggest, and this is apparently a difficulty that must be dealt with, see this paper.

In general, because the mass difference between the proton and neutron is about an MeV, in any system including protons and electrons at a temperature of order an MeV or higher, there will necessarily be populations of both neutrons and protons connected to each other by such processes, with relative amounts determined by the relevant Boltzmann factors. This should include systems where thermal fusion is taking place.

However the actual process of producing helium from hydrogen, as far as I understand, does not depend on capturing an electron on a proton to form a neutron. In stellar nucleosynthesis, two protons merge to form deuterium. That is, in the process of merging, one proton is converted into a neutron by the emission of a positron and a neutrino. Helium-2 (two protons) is highly unstable, so that this proton-to-neutron conversion producing stable deuterium is more important.

  • $\begingroup$ An electron has a rest-mass of about half a MeV, so wouldn't a temperature of half a MeV be sufficient? (of course, temperature is a statistical average, so it can still happen below that temperature, but more slowly) $\endgroup$
    – AI0867
    Jan 11, 2021 at 17:56
  • $\begingroup$ @AI0867 yeah I guess so, hence the word "order" ;) $\endgroup$
    – kaylimekay
    Jan 11, 2021 at 17:59
  • $\begingroup$ Yes, I got the "order" bit, but your phrasing entirely neglected the mass of the electron, which is quite relevant. $\endgroup$
    – AI0867
    Jan 11, 2021 at 18:01
  • $\begingroup$ The mass difference is about 1.3Mev, so you need a 800Kev electron. $\endgroup$ Jul 17 at 7:12

The neutron decay is time-invariant, so if one has

$$n \rightarrow p + e^- + \bar{\nu}$$

also the inverse process

$$e^- + \bar{\nu} + p \rightarrow n$$

is possible. Actually the in-going anti-neutrino is not necessary, because due to crossing-symmetry also the process

$$e^- + p \rightarrow n + \nu $$

is possible. The only thing one has to assure is that the electron-energy is in the right range, i.e. having an energy of a couple of MeV, in order to compensate for the mass difference between the proton that is lighter by about 1.3MeV than the neutron and to provide some energy (actually only little energy is necessary) to the now out-going neutrino. The electron-energy should not be excessively high in order to avoid deep-inelastic scattering.

The question how to effectively produce neutrons has to be posed. Shooting electrons on hydrogen-atoms is rather inefficient, as most electrons will not really come close to the proton/nucleus. Shooting electrons on uranium nuclei is more efficient as the nucleus already contains 92 protons. However, other problems might arise (radioactivity).


To answer the question of whether this occurs naturally: Yes. This is what happens with neutron degeneracy in neutron stars:

Neutron degeneracy is analogous to electron degeneracy and is demonstrated in neutron stars, which are partially supported by the pressure from a degenerate neutron gas. The collapse happens when the core of a white dwarf exceeds approximately 1.4 solar masses, which is the Chandrasekhar limit, above which the collapse is not halted by the pressure of degenerate electrons. As the star collapses, the Fermi energy of the electrons increases to the point where it is energetically favorable for them to combine with protons to produce neutrons (via inverse beta decay, also termed electron capture). The result is an extremely compact star composed of nuclear matter, which is predominantly a degenerate neutron gas, sometimes called neutronium, with a small admixture of degenerate proton and electron gases.


You are right, that a neutron is not just simply the "merge" of electron and proton, but there is such process when an electron and a proton form a neutron. For example it is a possible way of decay for proton rich atoms, and is called (K-)electron capture in that case.

But for me it seems a very rare process to create neutron in electron-proton collision by shooting electrons into hidrogen gas/plasma. I think the electrons and protons should be very close to eachother for a "long" time, what is not the case in this setup, but I'm not completly sure.

In fusion the source of neutrons is iether the decaying of protons (in proton rich nuclei) into neutrons and positrons (positive beta decay) or proton-proton "collisions" producing deuterium.

  • $\begingroup$ Protons decay into neutrons?? How does that work? $\endgroup$
    – releseabe
    Jan 10, 2021 at 13:51
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    $\begingroup$ @releseabe If there are too many protons in a nuclei, it can be energetically more favorable to "convert" one proton into a neutron. (Depending on the nuclei, it can be either a positive beta decay or a K-electron capture.) But protons can only decay into neutron inside a nucleus, a free proton will never decay into a neutron. $\endgroup$
    – fanyul
    Jan 10, 2021 at 14:14
  • $\begingroup$ @fanyul how absolutely certain is "never"? $\endgroup$
    – uhoh
    Jan 12, 2021 at 3:33
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    $\begingroup$ @uhoh As the neutron is heavier (with about 1.3 MeV$/c^2$), the proton should gain about 2 MeV energy (at least) to create a neutron and a pozitron. $\endgroup$
    – fanyul
    Jan 12, 2021 at 8:48
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    $\begingroup$ @uhoh Ahh, nwm. :) Actually it never decays into anything, there is a theorical half-life of about $10^{34}$ years. $\endgroup$
    – fanyul
    Jan 12, 2021 at 10:11

Electron capture decay can be induced, as well. It is now known that bound electron ground-state orbital radius is sustained by energy from the non-zero expectation value of the electronic component of the quantum vacuum as the bound electron emits Larmor radiation (a point change undergoing acceleration in relation to its electric field will emit Larmor radiation in the form of virtual photons, which is why every single element exhibits some form of magnetism, usually diamagnetism although certain valence electron configurations allow ferromagnetism to override the underlying diamagnetism)... at ground state, the energy emitted in the form of virtual photons exactly equals the energy received from the quantum vacuum.

Suppress that value (as, for instance, in a Casimir cavity) and you cause bound electron orbital radius to decrease and thus increase the probability of electron capture decay occurring. That transmutes the element up the Periodic Table under a sustained lower quantum vacuum energy density environment.

This has been (partially) done in laboratory... researchers were able to induce an artificial Lamb shift, causing the bound electrons of a noble gas to emit photons as the electrons quantum jumped to a lower orbital radius. Upon exiting the Casimir cavity, the ambient quantum vacuum restored bound electron orbital radius.

But the quantum vacuum is 'DC to daylight', so to speak, so it's exceptionally difficult to completely block it so as to induce rapid electron capture decay and thus transmutation up the Periodic Table.

[1] https://journals.aps.org/prd/abstract/10.1103/PhysRevD.11.790

[2] https://web.archive.org/web/20190713220130/https://arxiv.org/ftp/quant-ph/papers/0106/0106097.pdf

[3] https://web.archive.org/web/20190713225420/https://www.researchgate.net/publication/13330878_Ground_state_of_hydrogen_as_a_zero-point-fluctuation- determined_state "We show here that, within the stochastic electrodynamic formulation and at the level of Bohr theory, the ground state of the hydrogen atom can be precisely defined as resulting from a dynamic equilibrium between radiation emitted due to acceleration of the electron in its ground-state orbit and radiation absorbed from zero-point fluctuations of the background vacuum electromagnetic field, thereby resolving the issue of radiative collapse of the Bohr atom."

[4] https://web.archive.org/web/20180719194558/https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20150006842.pdf "The energy level of the electron is a function of its potential energy and kinetic energy. Does this mean that the energy of the quantum vacuum integral needs to be added to the treatment of the captured electron as another potential function, or is the energy of the quantum vacuum somehow responsible for establishing the energy level of the 'orbiting' electron? The only view to take that adheres to the observations would be the latter perspective, as the former perspective would make predictions that do not agree with observation."


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